b2 − 4ac reveals everything about a quadratic's roots before you solve. Every exam type — classifying roots, rational vs irrational, and finding k for equal / real / non-real roots — worked in full, with a worksheet.
For a quadratic ax2 + bx + c = 0, the part of the quadratic formula under the square root, b2 − 4ac, is the discriminant (symbol Δ). It tells you what kind of roots the equation has — without doing the full working. This guide covers every discriminant question type.
1What the discriminant tells you
- Δ > 0 : two different real roots (the graph crosses the x-axis twice).
- Δ = 0 : one real root, repeated (the graph just touches the x-axis).
- Δ < 0 : no real roots (the graph never reaches the x-axis).
- Δ > 0 and a perfect square (0, 1, 4, 9, 16…) : roots are rational and it factorises.
- Δ > 0 but NOT a perfect square : roots are real but irrational (surds).
2Type 1: Classify the roots (no real roots)
- 1Identify a = 2, b = −3, c = 5.Match to ax2 + bx + c = 0.
- 2Δ = b2 − 4ac = (−3)2 − 4(2)(5) = 9 − 40 = −31.Mind the sign: (−3)2 = +9.
- 3Δ = −31 < 0 → no real roots.A negative under the root has no real value.
3Type 2: Rational vs irrational roots
- 1Δ = (−5)2 − 4(1)(6) = 25 − 24 = 1.Compute the discriminant.
- 2Δ = 1 > 0 and 1 is a perfect square → two rational roots.Perfect-square Δ means it factorises.
- 1Δ = (−3)2 − 4(1)(1) = 9 − 4 = 5.
- 2Δ = 5 > 0 but not a perfect square → two real, irrational roots.The answer will contain 5.
4Type 3: Equal roots: find k
The common exam twist: an equation contains an unknown (often k) and you must find the value that gives a particular kind of root. Translate the words into a condition on Δ.
- 1Equal roots means Δ = 0.One repeated root ⇒ discriminant is zero.
- 2Δ = k2 − 4(1)(9) = k2 − 36 = 0.Substitute a = 1, b = k, c = 9.
- 3k2 = 36 → k = ±6.Two values give equal roots.
5Type 4: Real roots / non-real roots: find k
- 1Two real roots means Δ > 0.Real and distinct ⇒ discriminant positive.
- 2Δ = (−4)2 − 4(1)(k) = 16 − 4k > 0.Set up the inequality.
- 316 > 4k → k < 4.Solve the inequality (divide by 4).
Translate the words into a condition first: 'equal/one repeated root' → Δ = 0; 'two real roots' → Δ > 0; 'no real roots' → Δ < 0; 'rational roots' → Δ a perfect square. Then substitute and solve.
Sign of b2: if b = −3 then b2 = 9, not −9. Squaring always gives a positive — this single slip flips many answers.
Because the conditions are fixed and the algebra is short, discriminant questions are among the best-value marks in Grade 11–12 Paper 1. Drill every type below.
Practice exercises
Work each one out, then click to reveal the answer.
- 1Δ for x2 − 6x + 9 = 0, and the nature of roots. (classify)
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36 − 36 = 0 → one real (equal) root - 2Δ for x2 + 2x + 5 = 0, and the nature of roots. (classify)
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4 − 20 = −16 < 0 → no real roots - 3Δ for 2x2 − 7x + 3 = 0, and the nature. (classify)
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49 − 24 = 25 > 0 → two real roots - 4Are the roots of x2 − 4x + 4 = 0 rational? (rational?)
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Δ = 0 (perfect square) → rational (equal) - 5Rational or irrational: x2 − 2x − 2 = 0? (rational?)
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Δ = 4 + 8 = 12, not a perfect square → irrational - 6For which k does x2 + kx + 4 = 0 have equal roots? (equal, find k)
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k2 − 16 = 0 → k = ±4 - 7For which k does x2 − 6x + k = 0 have equal roots? (equal, find k)
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36 − 4k = 0 → k = 9 - 8For which k does x2 + 2x + k = 0 have two real roots? (real, find k)
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4 − 4k > 0 → k < 1 - 9For which k does x2 − 2x + k = 0 have no real roots? (non-real, find k)
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4 − 4k < 0 → k > 1 - 10Δ for 3x2 − 5x + 1 = 0. (compute)
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25 − 12 = 13 (two real, irrational)
Now practise it
Download Grade 11 past papers and worksheets on this topic.